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Laboratory Report Materials Chemistry Laboratory

The Kinetics Of The Reaction H₂O₂ + 2HI = 2H₂O +2I in Aqueous Solution Yufei Chang • Group X5


Abstract The aim of this experiment is to find out the rate constant and the energy of activation of the reaction between hydrogen peroxide and hydrogen iodide. The rate constant are measured at three different temperature ( 0℃ 20℃ 30℃) via the volumetric titration method. It’s obvious to find in the experiment temperature is proportional to the rate constant. As temperature increased , more kinetic energy supplied to overcome the activation energy barrier and the reaction becomes faster.

Introduction This reaction is kinetically of second order, not as be expected, third order. The mechanism is probably an initial, rate-determining step according to the equation H₂O₂ + I¯ → H₂O + IO¯ (slow), followed by a rapid reaction given by ) IO- + 2H+ + I → H2O + I2 (faster). ( The latter presumably occurs by consecutive steps, not by a simultaneous collision of four ions. The order of the reaction with respect to H₂O₂ can be studied conveniently by choosing condition such that there is practically constant excess of HI . The kinetics then follow the first order law . Experimentally, this is achieve by continually adding small volumes of sodium thiosulphate solution to remove the iodine as soon as it is liberated and to regenerate iodide according to the reaction 2Na₂S₂O₃+I₂ =Na₂S₄O₆+2NaI. By using a large volume of solution and adding small amounts of concentrated thiosulphate solution from a microburette, one can neglect the small increase of volume of the solution and take the concentration of I¯ ions as constant. The rate of the reaction then depends only on [H₂O₂] and temperature. The rate of change of concentration of A: d[A]/dt = k. [A]n [A] is the concentration of A at time t t is the time k is the rate constant n is the order of reaction If the reaction is pseudo first order with respect to H2O2:

or

t = elapsed time a = initial concentration of H2O2 a-x = concentration of H2O2 at time t


k = reaction rate constant c = constant

The rate constant would be obtained by using the equation above. After obtaining the rate constant at different temperatures, we used the Arrhenius equation to determine the activation energy (Potential Energy Barrier):

or

or

k = rate constant of the reaction A = a constant E= activation energy of reaction T = temperature in Kelvin

Experiment The Procedure should follow the method in Materials Chemistry Lab Script and handouts

Observations By timing the appearance of iodine (indicated by starch solution) after a small know volume of thiosulphate solution has been added. Another addition of thiosulphate is made immediately the blue colour reappears.

Result The vol.M/50 thiosulphate is 53.7 ml The volume of thiosulphate required to remove all the iodine produced in the reaction is expressed as a. Therefore, when 10ml of H2O2 was added into the flask, it would require 107.4ml of thiosulphate solution to complete the reaction. a= 107.4 ml at 0℃ Time of end point (sec)

Volume (ml)

a-x (ml)

log(a-x) (ml)

0

0.5

106.9

2.029

74

3.4

104

2.017

132

4.6

102.8

2.012

190

6

101.4

2.01


Time of end point (sec)

Volume (ml)

a-x (ml)

log(a-x) (ml)

247

7.3

100.1

2

300

9.1

98.3

1.99

377

10.4

97

1.987

449

11.8

95.6

1.98

511

14.2

93.2

1.969

Time of end point (sec)

Volume (ml)

a-x (ml)

log(a-x) (ml)

at 20℃

0

14.5

92.9

1.968

30

19.5

87.9

1.944

65

21.2

86.2

1.936

79

23.1

84.3

1.926

100

27.8

79.6

1.901

160

31.9

75.5

1.878

211

34.6

72.8

1.862

253

37.2

70.2

1.846

288

39.9

67.5

1.829

at 30℃ Time of end point (sec)

Volume (ml)

a-x (ml)

log(a-x) (ml)

0

5.7

101.7

2.007

35

11.3

96.1

1.982

65

15.2

92.2

1.964

88

17.8

89.6

1.952

105

20.4

87

1.941

122

23.2

84.2

1.925

145

25.3

82.1

1.914


Time of end point (sec)

Volume (ml)

a-x (ml)

log(a-x) (ml)

165

27.6

79.8

1.902

175

29.9

77.5

1.889


The gradient of the lines is equal to the rate constant K As shown on the graph At 0℃, K=-0.0001 At 20℃, K= -0.0005 At 30℃, K=-0.0008 Table-4

Temperature K

1/T (1/K)

K (1/s)

ln K

273

0.00366

0.0001

-9.21034

293

0.00341

0.0005

-7.6009

303

0.0033

0.0008

-7.1309


From the equation

Energy of the activation is 5889.6x8.31=48942.567 is 48.94 KJ/mol

Discussion The activation energy is the minimum amount of energy needed for colliding species to react. Usually one can think of the activation energy as the height of the potential barrier separating two minima of potential energy (of the reactants and of the products of reaction). For a chemical reaction to have noticeable rate, there should be noticeable number of molecules with the energy equal or greater than the activation energy.

Even if a sample has an evenly distributed temperature, their molecules do not all possess the same kinetic energy. The speed and thus the kinetic energy of the molecules are widely scattered. The distribution of the molecules' kinetic energy at a given temperature was mathematically described by Boltzmann (Boltzmann distribution law). In the illustration below, the Boltzmann distribution is depicted according to two different temperatures, whereat T1 < T2. As the T2 is higher ,larger fraction of the molecules has sufficient kinetic energy to overcome the barrier energy( EA) ,That is the reason why at higher temperature ,the rate constant is large.*

In another words, even a small increase in temperature causes a relatively large increase in the reaction rate. This fact is mathematically described by the Arrhenius equation, which


explains the influence of the temperature and the activation energy Ea on the rate constant k and thus on the reaction rate: k= k0 e-A/T, with A = Ea/R. Due to their definition, Ea and T (absolute temperature) are always positive. Therefore, according to the Arrhenius equation, the reaction rate is the higher the lower the activation energy Ea is, as Ea is part of the negative exponent's numerator. In contrast, the reaction rate increases accordingly when the temperature is raised, as T is part of the negative exponent's denominator. The error in this experiment is a bit large and this is due to these reactants were loss in transfers ,the effect of the external temperature and apparatus problem. From the result above could prove that the approximation of that the reaction rate doubled every 12 â&#x201E;&#x192; is acceptable. Reference Findlayâ&#x20AC;&#x2122;s practical physical chemistry by B.P Levitt Material science engineering by William Claysiter


Lab report the kinetics of the reaction